2-state system (an invert problem)

Given 2 states with energy $E_1 < E_2$, and interaction energy between the two states $V$, the Hamiltonian of the system is

$\displaystyle H = \begin{pmatrix} E_1 && V \\ V && E_2 \end{pmatrix}$

This system is already discussed in this post. The solution state in here again as:

The eigen-energies

$\epsilon_{\pm} = \bar{E} \pm \sqrt{dE^2 + V^2 }$

the eigen-states are

$\Phi_{+} = \alpha \phi_1 + \beta \phi_2$

$\Phi_{-} = -\beta \phi_1 + \alpha \phi_2$

where

$\displaystyle \alpha = \frac{dE + \sqrt{dE^2 + V^2 }}{\sqrt{(dE+\sqrt{dE^2+V^2})^2 + V^2}}, \beta = \frac{V}{\sqrt{(dE+\sqrt{dE^2+V^2})^2 + V^2}}$

The $\phi_{1,2}$ are the wavefunction of pure state of $E_{1,2}$.

This can be easily extended to 3-state system or $n$-state system. The number of coupling constants is $n(n-1)/2$.

Experimentally, we observed the mixed states and the spectroscopic factors originated from one of the pure state. For example, in a neutron transfer reaction, the transferred neutron may coupled with the 2+ state of the core. The neutron is sitting in a pure orbital ($\phi_1$), say 1d5/2 orbital, it couple with the 2+ state ($\phi_2$) with interaction energy $V$. The final states will have energy $\epsilon_{\pm}$ and we extract the spectroscopic factors ($\alpha^2, \beta^2$) for the 1d5/2 orbital \$.

To state more clear, in a transfer reaction, we may observed two excited states $A_+, A_-$ with transfer of $|0p_{1/2}\rangle$ neutron. This neutron assumed to couple with a core state $|C\rangle$ and a core excited state $|C^+\rangle$. And we supposed that the 2 observed states are:

$|A_+ \rangle = \alpha |0p_{1/2}\rangle |C\rangle + \beta |0p_{1/2} \rangle |C^+\rangle = \alpha |\phi_1\rangle + \beta |\phi_2\rangle$

$|A_- \rangle = \beta |0p_{1/2}\rangle |C\rangle - \alpha |0p_{1/2} \rangle |C_+\rangle = \beta |\phi_1\rangle - \alpha |\phi_2\rangle$

We then want to find out the un-perturbed energy $E_{1,2}$ and the interaction $V$. And this is actually an easy problem by the diagonalization process.

The Hamiltonian was diagonalized into

$\displaystyle H = \begin{pmatrix} E_1 && V \\ V && E_2 \end{pmatrix} = P \cdot D \cdot P^{-1}$

where

$\displaystyle P = \begin{pmatrix} \alpha && -\beta \\ \beta && \alpha \end{pmatrix} , det(P) = 1$

$\displaystyle D = \begin{pmatrix} \epsilon_+ && 0 \\ 0 && \epsilon_- \end{pmatrix}$

Thus,

$E_1 = \alpha^2 \epsilon_+ +\beta^2 \epsilon_-$

$E_2 = \beta^2 \epsilon_+ +\alpha^2 \epsilon_-$

$\displaystyle V = \alpha \beta (\epsilon_+ - \epsilon_-) = \frac{1}{2}\sqrt{d\epsilon^2 - dE^2}$

Spectroscopic factor & Occupation number

Started from independent particle model, the Hamiltonian of a nucleus with mass number $A$ is

$H_A = \sum\limits_{i}^{A} h_i + \sum\limits_{i, j>i} V_{ij}$

we can rewrite the Hamiltonian by isolating a nucleon

$H_A = H_B + h_1 + V_{B1}$

than, we can use the basis of $H_B$ and $h_1$ to construct the wavefunction of the nucleus A as

$\left|\Phi_A\right>_{J} = \sum\limits_{i,j} \beta_{ij} \left[ \left|\phi_i\right>\left|\Phi_{B}\right>_{j} \right]_{J}$

where the square bracket is anti-symmetric angular coupling between single particle wavefunction $\left|\phi_i\right>$ and wavefunction $\left|\Phi_B\right>_{j}$. The $\beta_{ij}$ is the spectroscopic amplitude.

The square of the spectroscopic amplitude times number of particle $n_i$ at state $\left|\phi_i\right>$ is the spectroscopic factor of the nucleon at state $\left|\phi_i\right>$ and nucleus B at state $\left| \Phi_B\right>_j$

$S_{ij} = n_i \beta_{ij}^2$

The occupation number is the sum of the spectroscopic factors of the nucleus B

$\sum\limits_{j} S_{ij} = n_i$

After the definition, we can see, when the nucleon-core interaction $V_{B1}$ is neglected, the spectroscopic factor is 1 and the occupation number is also 1.

Since the $\beta$ is coefficient for changing basis from $\left|\Phi_A\right>_J$ into basis of $\left[ \left|\phi_i\right>\left|\Phi_{B}\right>_{j} \right]_{J}$ Thus, the matrix $\beta$ is unitary

$\beta\cdot \beta^\dagger = 1$

thus, each column of row vector of $\beta$ is normalized.

The properties of $\beta$ can be found by solving the eigen system of the $H_A$ from the core $H_C = H_B + h_1$. The core hamiltonian is diagonal. The nucleon-core interaction introduce diagonal terms and off-diagonal terms. When only diagonal terms  or monopole term exist, only the eigen energy changes but the eignestate unchanges. Therefore, the configuration mixing is due to the off-diagonal terms.

However, when there are off-diagonal terms, the change of diagonal terms will changes the mixing.

In degenerate 2 states system,  the Hamiltonian be

$H= \begin{pmatrix} E_1 & V \\ V & E_2 \end{pmatrix}$

The eigen energy are $\bar{E} \pm \sqrt{ dE^2 + V^2}$, where $\bar{E} = (E_1+E_2)/2$ and $dE = (E_1 - E_2)/2$, The eigen vector are

$\displaystyle \vec{v}_{\pm} = \frac{1}{\sqrt{(dE \pm \sqrt{dE^2+V^2})^2 +V^2}} ( dE \pm \sqrt{dE^2 +V^2}, V)$

Since the eigenvector must be orthogonal. If we set one eigenvector be

$\displaystyle \vec{v}_+ = { \alpha, \beta}, \alpha^2 + \beta^2 = 1$

The other eigen vector must be

$\displaystyle \vec{v}_- = { -\beta, \alpha}$

One can check that

$\displaystyle \frac{V}{\sqrt{(dE + \sqrt{dE^2+V^2})^2 +V^2}} = \frac{ dE - \sqrt{dE^2 +V^2} }{\sqrt{(dE - \sqrt{dE^2+V^2})^2 +V^2}}$

When the states are degenerated, $E_1 = E_2 = E$, the eigen energy is $E \pm V$, and the eigen state is $\frac{1}{\sqrt{2}} (1,1)$

As we can see, the eigen state only depends on the difference of the energy level, thus, we can always subtract the core energy and only focus on a single shell. For example, when we consider 18O, we can subtract the 16O binding energy.

In above figure, we fixed the $E_1$ and $E_2$. We can see the spectroscopic factor decreases for the lower energy state (red line), and the state mixing increase.

We now fixed the $V$ and $E2$. When $E_1 < E_2$, the particle stays more on the $E_1$ state, which is lower energy. The below plot is the eigen energy. Even though the mixing is mixed more on the excited state, but the eigen energy did not cross.

Nuclear correlation & Spectroscopic factor

In the fundamental, correlation between two objects is

$P(x,y) \neq P(x) P(y)$

where $P$ is some kind of function. To apply this concept on nuclear physics, lets take a sample from 18O. 18O can be treated as 16O + n + n. In the independent particle model (IPM), the wave function can be expressed as

$\left|^{18}O\right>= \left|^{16}O\right>\left|\phi_a\right>\left|\phi_b\right>= \left|^{16}O\right>\left|2n\right>$

where the wave function of the two neutrons is expressed as a direct product of two IPM eigen wave functions, that they are un-correlated. Note that the anti-symmetry should be taken in to account, but neglected for simplicity.

We knew that IPM is not complete, the residual interaction has to be accounted. According to B.A. Brown, Lecture Notes in Nuclear Structure Physics [2011], Chapter 22, the 1s1/2 state have to be considered. Since the ground state spins of 18O and 16O are 0, thus, the wavefunction of the two neutrons has to be spin 0, so that only both are in 1d5/2 or 2s1/2 orbit. Thus, the two neutrons wave function is

$\left| 2n \right> = \alpha \left|\phi_1\right>\left|\phi_1\right>+\beta \left|\phi_2\right>\left|\phi_2\right>$

when either $\alpha$ or $\beta$ not equal 0, thus, the two neutrons are correlation. In fact, the $\alpha = 0.87$ and $\beta = 0.49$.

The spectroscopic factor of the sd-shell neutron is the coefficient of $\alpha$ times a isospin-coupling factor.

From the above example, the correlation is caused by the off-diagonal part of the residual interaction. To be more specific, lets take 18O as an example. The total Hamiltonian is

$H_{18} = H_{16} + h_1 + h_2 + V$

where $h_1 = h_2 = h$ is the mean field or single particle Hamiltonian

$h\left|\phi_i\right>= \epsilon_i\left|\phi_i\right>$

since $H_{16}$ is diagonal and not excited (if it excited, then it is called core polarization in shell model calculation, because the model space did not included 16O.), i.e. $H_{16} = \epsilon_{16} I$, we can neglect it in the diagonalization of the $h_1+h_2 + V$ and add back at the end. In the 1d5/2 and 2s1/2 model space, in order to form spin 0, there is only 2 basis,

$\left|\psi_1\right> = \left|\phi_1\right>\left|\phi_1\right>$ and

$\left|\psi_2\right> = \left|\phi_2\right>\left|\phi_2\right>$

express the Hamiltonian in these basis,

$V = \begin{pmatrix} -1.79 & -0.83 \\ -0.83 & -2.53\end{pmatrix}$

Because of the diagonalization, the two states $\left|\psi_1\right>$ and$\left|\psi_2\right>$ are mixed, than the two neutrons are correlated. This is called configuration mixing.

According to B.A. Brown,

the configuration mixing on the above is long-ranged correlation (LRC). It is near the Fermi surface and the energy is up to 10 MeV.

The short-ranged correlation (SRC) is caused by the nuclear hard core that scattered a nucleon to highly single particle orbit up to 100 MeV.

The LRC is included in the two-body-matrix element. The SRC is included implicitly through re-normalization of the model space.

There is a correlation due to tensor force. Since the tensor force is also short-ranged, sometimes it is not clear what SRC is referring from the context. And the tensor force is responsible for the isoscalar pairing.

The discrepancy of the experimental spectroscopic factor and the shall model calculation is mainly caused by the SRC.